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Problem 1162

AIME late

Algebra Difficulty 5.1 Prove it THE 68th ROMANIAN MATHEMATICAL OLYMPIAD · Romania

Let $f$ and $g$ be continuous real-valued functions on the closed unit interval $[0, 1]$ such that $f(x)g(x) \ge 4x^2$ for all $x$ in $[0, 1]$ . Show that (at least) one of the integrals
$\int_{0}^{1} f(x) \, dx, \quad \int_{0}^{1} g(x) \, dx$
has an absolute value greater than or equal to $1$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

The functions $f$ and $g$ vanish at no point in the half-open interval $(0, 1]$ , so $\left|\int_{0}^{1} f(x) \, dx\right| = \int_{0}^{1} |f(x)| \, dx$ and $\left|\int_{0}^{1} g(x) \, dx\right| = \int_{0}^{1} |g(x)| \, dx$ . Consequently,
$\begin{aligned} 1 &= \int_{0}^{1} 2x \, dx \le \int_{0}^{1} \sqrt{f(x)g(x)} \, dx = \int_{0}^{1} \sqrt{|f(x)||g(x)|} \, dx \\ &\le \int_{0}^{1} \frac{|f(x)| + |g(x)|}{2} \, dx = \frac{1}{2} \left( \int_{0}^{1} |f(x)| \, dx + \int_{0}^{1} |g(x)| \, dx \right), \end{aligned}$
and the conclusion follows.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.